Question

18-22. In analogy to the characteristic vibrational temperature, we can define a characteristic electronic temperature by @ele, s = where is the energy of the jth excited electronic state relative to the ground state. Show that if we define the ground state to be the zero of energy, then Telee = 8+8, e-Ociec.1/7 +8,e-Pelec.2/T +... The first and second excited electronic states of O(g) lie 158.2 cm-' and 226.5 cm- above the ground electronic state. Given 8, = 5,8, = 3, and g, = 1, calculate the values of elec.], elec. 2, and Celec (ignoring any higher states) for O(g) at 5000 K.

Answer 1

Ane: I first of all we have to calculate the energy of bisst s second excited electroner state. alow for 1st excited electronic state. . Ñ (ware number) = 158.2cm Een her -34 10 t = 6.62x10 J5 x 3 x 10 cms x 158.2 cm -24 0.87 X 10 J @eleri - Fer 1 Again -24 frei 301.87 x10 3 = 218.75 X10 K KB 1.38×10 23 JK' = 21.875k for and executed electron state , 9 = 226.5em fea - her -34 = 6.62x10 J5 x 3x10 cm 5'x 226.5cm = 4498.29 x10-24 cles - Fez , 4498.29 * 10° ſ = 3259.63x10k - 1.38 x 10 JK = 325.963k

Now elec go tgpe o lift a gre Beliz/ 5+32 21.875/5000tie 395-967/s000 =5+32 0.00437 -0.0651 te = 5+3 *.9957) + (93697) - 8.9240 elen = 21.875K Deler 2 = 323.963k Tele = 8.9240